Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <queue>
#include <vector>
#include <iostream>
#include <functional>

#pragma warning(disable : 4996)

#define INF (1 << 23)

using namespace std;

int main()
{
	int N, M, s, t, u, v;

	scanf("%d", &N);
	scanf("%d", &M);
	scanf("%d", &s); s--;
	scanf("%d", &t); t--;

	vector<vector<int> > G(N);

	for (int i = 0; i < M; i++)
	{
		scanf("%d", &u);
		scanf("%d", &v);

		G[u - 1].push_back(v - 1);
		G[v - 1].push_back(u - 1);
	}

	vector<int> d1(N, INF); d1[s] = 0;
	vector<int> d2(N, INF); d2[t] = 0;

	queue<int> que;

	que.push(s);

	while (!que.empty())
	{
		int node = que.front(); que.pop();

		for (int i = 0; i < G[node].size(); i++)
		{
			if (d1[G[node][i]] == INF)
			{
				d1[G[node][i]] = d1[node] + 1; que.push(G[node][i]);
			}
		}
	}

	que.push(t);

	while (!que.empty())
	{
		int node = que.front(); que.pop();

		for (int i = 0; i < G[node].size(); i++)
		{
			if (d2[G[node][i]] == INF)
			{
				d2[G[node][i]] = d2[node] + 1; que.push(G[node][i]);
			}
		}
	}

	int shortest = d1[t];

	vector<int> r1(shortest - 1, 0); for (int i = 0; i < N; i++) { if (d1[i] < shortest - 1) { r1[d1[i]]++; } }
	vector<int> r2(shortest - 1, 0); for (int i = 0; i < N; i++) { if (d2[i] < shortest - 1) { r2[d2[i]]++; } }

	long long ret = 0;

	for (int i = 0; i < shortest - 1; i++)
	{
		ret += 1LL * r1[i] * r2[shortest - i - 2];
	}
	
	printf("%lld\n", ret);

	return 0;
}